605 Views

A steady current element of 10e-3 az A/m is located at the origin in free space. Find the magnetic field intensity dH at a point (1,0,0)


Rishi Rishi 11th Dec, 2020
Answer (1)
Ayush 19th Mar, 2021

Hello candidate,

The magnitude of Magnetic field intensity refers to the number of ampere turns which the the electric or the the conducting coil makes on a unit length to produce a magnetic effect on a solenoid or any other magnetic body.

Hence, the value of magnetic field intensity is equal to current element multiplied by the space coordinates, which gives 10e^-3 *(0+0+1)= 10e^-3 H.

Hope it was helpful!!

Related Questions

UPES B.Tech Admissions 2026
Apply
Ranked #43 among Engineering colleges in India by NIRF | Highest Package 1.3 CR , 100% Placements
UPES Integrated LLB Admission...
Apply
Ranked #18 amongst Institutions in India by NIRF | Ranked #1 in India for Academic Reputation by QS Rankings | 16 LPA Highest CTC
Nirma University Law Admissio...
Apply
Grade 'A+' accredited by NAAC | Ranked 33rd by NIRF 2025
UPES M.Tech Admissions 2026
Apply
Ranked #45 Among Universities in India by NIRF | 1950+ Students Placed 91% Placement, 800+ Recruiters
UPES | BBA Admissions 2026
Apply
#36 in NIRF, NAAC ‘A’ Grade | 100% Placement, up to 30% meritorious scholarships
IMT Ghaziabad PGDM Admissions...
Apply
AACSB, NBA & SAQS Accredited | H-CTC 41.55 LPA | Merit Based Scholarship
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books